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Let $G$ be a discrete countable group acting on a compact, Hausdorff space $X$. Assume that the action is transitive. Namely, $G\cdot x=X$, for all $x\in X$.

Does it follow that $X$ is finite?

I thought that the answer should be yes, as $G\cdot x$ is then a discrete compact space, and so must be finite. However, I am not sure anymore that I can claim that it is a discrete space. I'd appreciate any help.

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    $\begingroup$ Notice that under these conditions $G/G_x\cong G\cdot x=X$ homeomorphism (see theorem 6.2 in the book "Crossed Products" by Williams). Thus $G/G_x$ is discrete and compact, so finite, and so is $X$. $\endgroup$ – Shirly Geffen Jun 23 '19 at 15:50
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    $\begingroup$ @ShirlyGeffen $G/G_x$ need not be homeomorphic to $Gx$. The standard map is a continuous bijection only. There are counterexamples when it is not a homeomorphism. It is true however when $G$ is compact for example (not our case). The theorem most likely assumes something more about $G$ or the action. $\endgroup$ – freakish Jun 23 '19 at 15:53
  • $\begingroup$ @freakish It is a homeomorphism if $G\cdot x$ is locally closed in $X$, which is the case here. $\endgroup$ – Shirly Geffen Jun 23 '19 at 15:55
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    $\begingroup$ @ShirlyGeffen the theorem also assumes that $X$ is second countable. Which isn't clear for me why that holds. And with that assumption this can be easily proved using Baire category theorem instead. $\endgroup$ – freakish Jun 23 '19 at 16:08
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Generally any topological space is an image of some discrete space. So your conclusion that $G\cdot x$ is discrete is not necessarily true, or requires deeper explanation. But in this scenario it works. Not because $G$ is discrete but because it is countable. First of all note that $X$ is also countable as an image of a countable set.

Assume that $X$ is not discrete. Then there is a point $x_0\in X$ which is not isolated. Since $G$ acts on $X$ transitively and $x\mapsto gx$ is a homeomorphism then this shows that no point in $X$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable. For the proof see here (plus some discussion regarding related set theoretic axioms, for safety I assume ZFC). Contradiction.

Since $X$ is discrete and compact then it has to be finite.

Note that the assumption about $G$ being discrete is irrelevant.

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    $\begingroup$ Actually $G$ being discrete is irrelevant but is no restriction since one can always consider $G$ with the discrete topology and proceed. $\endgroup$ – YCor Jun 24 '19 at 8:21
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Since $G$ is countable and $G\cdot x=X$, we have $X$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $X$ is homeomorphic to $\omega^\alpha\cdot n+1$, where $\alpha+1$ is the Cantor-Bendixon rank of $X$, and $n\geq 1$ is the cardinality of $X^{(\alpha)}$.

Now we need to rule out $\alpha\geq 1$. For such cases, note that $G$ can only map $\omega^\alpha\cdot m$, $m>0$ to another point $\omega^\alpha\cdot m'$, $m'>0$, because every neighbourhood of $\omega^\alpha\cdot m$ contains a homeomorphic copy of $\omega^\alpha+1$. So the action of the homeomorphism group (hence $G$) is not transitive unless $\alpha=0$.

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    $\begingroup$ Actually the overkill is based on studying isolated points and uses at the first place that a nonempty countable Hausdorff space has an isolated point, which is enough to proceed (as in freakish's answer). $\endgroup$ – YCor Jun 24 '19 at 8:23

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